composed variable after Object Operator in PHP

Question

How can I build a composed variable while creating a variable in PHP? (Sorry I'm not sure how to call the different elements)

This is what I'm trying to do:

$language = 'name_'.$this->session->userdata('site_lang');

for ($i=1;$i<=3;$i++) {
    $data = $arraydata->$language_.$i; // problem is here
}

I would like $language_.$i to be equivalent to name_english_1, next loop name_english_2... The same way I built $language

don't generate dynamic variable names. that's just bad coding. use an array. $arraydata->$language[$i] is FAR easier to maintain/understand. — Marc B
– Marc B, Commented Apr 29, 2016 at 20:58
but the fields in my database are called ´name_english_1´, ´name_english_2´ so how could I retrieve them? — remyremy
– remyremy, Commented Apr 29, 2016 at 21:01

Barmar · Accepted Answer · 2016-04-29 22:10:30Z

3

If you want to use an expression in a computed property, you have to put the expression in braces. Also, you need to put the underscore in quotes.

$data = $arraydata->{$language."_".$i};

However, I suggest you redesign your data structure. Instead of having separate name_LANG_i properties, make a single name property whose value is a multi-dimensional array.

$lang = $this->session->userdata('site_lang');

for ($i=1;$i<=3;$i++) {
    $data = $arraydata->name[$lang][$i];
    // do something with $data
}

Whenever you find yourself using variable variables or variable properties, it's almost always a sign that you should be using an array instead.

edited Apr 29, 2016 at 22:10

answered Apr 29, 2016 at 22:05

Barmar

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Comments

Rajdeep Paul · Accepted Answer · 2016-04-29 21:18:44Z

0

First construct the field name and then use it for accessing the field value from the object $arraydata. So your code should be like this:

$language = 'name_'.$this->session->userdata('site_lang');
for ($i = 1; $i <= 3; $i++) {
    $var = "{$language}_{$i}";
    $data = $arraydata->$var;

    // echo $data;

}

answered Apr 29, 2016 at 21:18

Rajdeep Paul

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